The product of three consecutive positive integers is divisible by 6' . Is this statement true or false? justify your answer

(i) False As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6} ⇒ {2, 3, 4, 5} ∩ {3, 6} = {3}  (ii) False As a ∈ {a, e, i,...

Solution: True. Justification: Let a, a + 1 be two consecutive positive integers. By Euclid’s division lemma, we have a = bq + r, where 0 ≤ r < b For b = 2 , we...

Solution: True. Justification: At least one out of every three consecutive positive integers is divisible by 2. Therefore, The product of three consecutive positive integers is divisible by 2. At least one out of every...

Solution: Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer. By Euclid’s division lemma, we have a = bq + r; 0 ≤...

Let the three consecutive integers bex, x+1and x+2. According to the question, x+x+1+x+2 = 51 => 3x+3=51 => 3x+3-3=51-3 [Subtracting 3 from both sides] => 3x=48 => 3x/3 = 48/3 [Dividing both sides by 3] X=16 Hence, first...

Let, (n – 1) and n be two consecutive positive integers ∴ Their product = n(n – 1) = n 2 −n We know that any positive integer is of the form 2q...

Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or, 6q + 3 or 6q +...

True,because n(n+1) will always be even, as one out of the n or (n+1) must be even.

(i) False As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6} ⇒ {2, 3, 4, 5} ∩ {3, 6} = {3} (ii) False As a ∈ {a, e, i, o, u}, a...

True, because n(n + 1) will always be even, as one out of the n or n+1 must be even.